Thread: Boolean algebra help

Hi,

I have a boolean function that im not sure how to solve.



im supposed to write this on disjunctive normal form.

I tried multiplying the and the term
by (z+z') and (y+y') but that makes the whole expression really long.

Any ideas?

Jones

I assume (x+z')' means not ( x or (not z)). If so, then doesn't this simplify to
x'z? Then your expression simplifies to:

= x'zy'z + xy' + xz'
= x'y'z + xy' + xz'

No, thats incorrect.

The correct answer should be

Note:
xy' = xy'z + xy'z'
because z + z'=1

So,


Simplifying the first two paranthesis


Should simplify down to just

Applying Morgans law gives:



Using de morgans again gives:

The whole expression now becomes

Is this correct?

you need my advice it is a 3 literals function so it is easy to build the truth table and see where the function set to 1 then build the karnough map and you will get the most simplified form

the simplified form is z'x+zy'

waiting your reply

truth table :
X Y Z F
0 0 0 0
0 0 1 1
0 1 0 0
0 1 1 0
1 0 0 1
1 0 1 1
1 1 0 1
1 1 1 0

K-map :
xy 00 01 11 10
z 0 0 0 (1 1)
1 1) 0 0 (1
=> z'x+zy'